Finding the fixed field corresponding to Galois group $\textrm {Gal} (\mathbb Q (\root p \of 2, \zeta_p)$ for prime $p$.

Question

I've already shown that this is the splitting field of the polynomial $x^p-2$ over $\mathbb Q$, and that the degree of extension of $L=\mathbb Q (\root p \of 2,\zeta_p)$ is $p(p-1)$. Let $H$ be the subgroup of $\textrm {Gal} (L/\mathbb Q)$ such that for all $\sigma \in H$, $\sigma (\root p \of 2)=\root p \of 2$. I wish to find the fixed field corresponding to $H$.
Since $L^H$ contains $\mathbb Q$ and $\root p \of 2$, by definition $\mathbb Q (\root p \of 2) \subseteq L^H$. I'm stuck trying to show inclusion in the other direction. I've tried looking at the degrees of the extensions but I can't conclude that $[L^H:\mathbb Q (\root p \of 2)]=1$.

Answer 1

The Galois group $G:=Gal(L/\mathbb{Q})$ has order $[L:\mathbb{Q}]=p(p-1)$. An automorphism $\sigma\in G$ of $L=\mathbb{Q}(\sqrt[p]{2},\zeta_p)$ is determined by $\sigma(\sqrt[p]{2})$ and $\sigma(\zeta_p)$. The automorphisms must send $\sqrt[p]{2}$ and $\zeta_p$ to roots of their respective minimal polynomial : $X^p-2$ and $\Phi_p=X^{p-1}+\ldots+X+1$. All the automorphisms are :

$$\bigg\{\sigma:\sqrt[p]{2}\mapsto \zeta_p^i\sqrt[p]{2} , \zeta_p\mapsto \zeta_p^j\big\lvert i\in \{0,1\ldots p-1\} \mbox { and } j\in\{1,\ldots p-1\}\bigg\}$$

As you see we have exactly $p(p-1)$ $\mathbb{Q}$-automorphisms of L. Let H be the subgroup such that for all $\sigma\in H$ we have $\sigma(\sqrt[p]{2})=\sqrt[p]{2}$. One inclusion is clear : $\mathbb{Q}(\sqrt[p]{2})\subset L^H$.

Consider now the following tower of extensions : $$\mathbb{Q}\subset \mathbb{Q}(\sqrt[p]{2})\subset L^H\subset L$$ The extension $L^H\subset L$ is galois with Galois group $H=Gal(L/L^H)$ and order $[L:L^H]=\#H$. It consists of all automorphisms which are the identity on $\mathbb{Q}(\sqrt[p]{2})$. Let $\sigma\in H$, then $\sigma:L\longrightarrow L$ and is determined by the images of $\sqrt[p]{2}$ and $\zeta_p$. We know that $\sigma(\sqrt[p]{2})=\sqrt[p]{2}$ and this leaves us with the following question: What are the possible images of $\zeta_p$ under this $\sigma$? The answer is exactly p-1. So we claim : $$H=\{\sigma:\sqrt[p]{2}\mapsto \sqrt[p]{2},\zeta_p\mapsto\zeta_p^i| i=1,\ldots p-1\}$$

The inclusion $\{\sigma:\sqrt[p]{2}\mapsto \sqrt[p]{2},\zeta_p\mapsto\zeta_p^i| i\in\{1,\ldots p-1\}\}\subset H$ is clear. For the other inclusion note that if H had more than these $p-1$ automorphisms, then necessarily it can't be the identity on $\mathbb{Q}(\sqrt[p]{2})$, it must send $\sqrt[p]{2}$ to $\zeta_p^i\sqrt[p]{2}$ for $i=1,2\ldots p-1$.

So H equals this set and has order $p-1$. From the multiplicativity rule for towers of field extensions it follows that : $[L^H:\mathbb{Q}]=p$. Together with the inclusion $\mathbb{Q}(\sqrt[p]{2}\subset L^H$ we now got the equality :

$$\mathbb{Q}(\sqrt[p]{2})=L^H$$